Question: A particle moves along the curve $xy=10$ so that the $y$ -coordinate is increasing at a constant rate of $3$ units per minute. What is the rate of change (in units per minute) of the particle's $x$ -coordinate when the particle is at the point $(2,5)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{5}{2}$ (Choice B) B $0$ (Choice C) C $-\dfrac{6}{5}$ (Choice D) D $\dfrac{4}{5}$
Answer: Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This, however, shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dy}{dt}=3$ for any value of $t$. We are asked for the rate of change of the particle's $x$ -coordinate when the particle is at the point $(2,5)$. In other words, we are asked for the value of $\dfrac{dx}{dt}$ at the point $(2,5)$. Finding $\dfrac{dx}{dt}$ $\dfrac{dx}{dt}=-\dfrac{3x}{y}$ Finding $\dfrac{dx}{dt}$ at $(2,5)$ The expression for $\dfrac{dx}{dt}$ depends on both the particle's $x$ -coordinate ${2}$ and its $y$ -coordinate ${5}$ : $\begin{aligned} \dfrac{dx}{dt}&=-\dfrac{3({2})}{({5})} \\\\ &=-\dfrac{6}{5} \end{aligned}$ In conclusion, the rate of change of the particle's $x$ -coordinate when the particle is at the point $(2,5)$ is $-\dfrac{6}{5}$ units per minute.